Download Estonian Math Competitions 2007/2008

Estonian Math Competitions
2007/2008
The Gifted and Talented Development Centre
Tartu 2008
WE
THANK:
Problem authors: Juhan Aru, Oleg Koshik, Toomas Krips, Kaie Kubjas, Härmel Nestra,
Hendrik Nigul, Uve Nummert, Reimo Palm, Oleksandr Rybak (Ukraine),
Laur Tooming, Jan Willemson, Indrek Zolk
Translation: Emilia Käsper, Ago-Erik Riet
Edition: Reimo Palm, Indrek Zolk
Estonian Mathematical Olympiad
Mathematics Contests in Estonia
The Estonian Mathematical Olympiad is held annually in three rounds – at the school,
town/regional and national levels. The best students of each round (except the final)
are invited to participate in the next round. Every year, about 110 students altogether
reach the final round.
In each round of the Olympiad, separate problem sets are given to the students of each
grade. Students of grade 9 to 12 compete in all rounds, students of grade 7 to 8 participate at school and regional levels only. Some towns, regions and schools also organise
olympiads for even younger students. The school round usually takes place in December, the regional round in January or February and the final round in March or April in
Tartu. The problems for every grade are usually in compliance with the school curriculum of that grade but, in the final round, also problems requiring additional knowledge
may be given.
The first problem solving contest in Estonia took place already in 1950. The next one,
which was held in 1954, is considered as the first Estonian Mathematical Olympiad.
Apart from the Olympiad, open contests are held twice a year, usually in October and
in December. In these contests, anybody who has never been enrolled in a university
or other higher education institution is allowed to participate. The contestants compete
in two separate categories: the Juniors and the Seniors. In the first category, students
up to the 10th grade are allowed to participate; the other category has no restriction.
Being successful in the open contests generally assumes knowledge outside the school
curriculum.
According to the results of all competitions during the year, about 20 IMO team candidates are selected. IMO team selection contest for them is held in April or May. This
contest lasts two days; each day, the contestants have 4.5 hours to solve 3 problems,
similar to the IMO. All participants are given the same problems. Some problems in our
selection contest are at the level of difficulty of the IMO but somewhat easier problems
are usually also included.
The problems of previous competitions can be downloaded from
http://www.math.olympiaadid.ut.ee/eng.
Besides the above-mentioned contests and the quiz “Kangaroo” some other regional
competitions and matches between schools are held as well.
*
This booklet contains problems that occurred in the open contests, the final round of
national olympiad and the team selection contest. For the open contests and the final
round, selection has been made to include only original and interesting problems. The
team selection contest, containing only original problems, is presented entirely.
1
Selected Problems from Open Contests
OC-1. An n-boomerang consists of 2n − 1 unit squares arranged in an
L-shape with both legs of length n (n = 4 in the figure). Find all integers
n > 2 for which there exists a rectangle with integer side lengths that can
be partitioned into n-boomerangs. (Juniors.)
Answer: the only suitable integer is 2.
Solution. If n = 2, the rectangle of size 2 × 3 can be partitioned into two 2-boomerangs
(Fig. 1).
Let us prove that if n > 3 then there are no rectangles that can be partitioned into
n-boomerangs. Let ( x, y) denote the unit square located in row x and column y, where
x and y are positive integers. Denote by ( a, b) − (c, d) − (e, f ) the boomerang with endsquares in unit squares ( a, b) and (e, f ) and corner-square in unit square (c, d). Clearly
(1, 1) has to be covered by an end-square or a corner-square of some boomerang.
If a boomerang covers (1, 1) with its corner-square then (2, 2) can be covered by another
boomerang again with a corner-square or an end-square.
Figure 1
Figure 2
Figure 3
Figure 4
1. If another boomerang covers (2, 2) with a corner-square (Fig. 2) then the
boomerangs covering (n + 1, 1) and (1, n + 1) leave the squares (n + 2, 2) and
(2, n + 2) empty. For any position of the boomerang covering (3, 3), one of (n + 2, 2)
and (2, n + 2) can not be covered anymore.
2. If another boomerang covers (2, 2) with its end-square (Fig. 3) then let it be without
loss of generality positioned as (2, 2) − (2, n + 1) − (n + 1, n + 1). The boomerang
covering (3, 2) is then positioned as (3, 2) − (n + 2, 2) − (n + 2, n + 1). The second
and third boomerang now have an isolated empty rectangle with side lengths less
than n between them.
If the first boomerang covers (1, 1) with its end-square (Fig. 4) then we can assume
without loss of generality that it is positioned as (1, 1) − (1, n) − (n, n). The boomerang
covering (2, 1) must now be positioned as (2, 1) − (n + 1, 1) − (n + 1, n). Those two
have an empty rectangle in between that is too small to fit any boomerangs.
OC-2. Do there exist four different integers a, b, c, d, all greater than one, satisfying
gcd( a, b) = gcd(c, d) and a) ab = cd; b) ac = bd? (Juniors.)
2
Answer: a) yes; b) no.
Solution 1. a) Let x, y, z and w be arbitrary different pairwise co-prime positive integers.
Let a = xy, b = zw, c = xz and d = yw. All these numbers are greater than 1. Then
gcd( a, b) = gcd( xy, zw) = 1 and also gcd(c, d) = 1, whereas ab = cd = xyzw.
b) Assume for contradiction that such a, b, c, d exist. Let s = gcd( a, b) = gcd(c, d). Write
a = a′ s, b = b′ s, c = c′ s, d = d′ s, then gcd( a′ , b′ ) = 1 and gcd(c′ , d′ ) = 1. The equation
ac = bd becomes a′ s · c′ s = b′ s · d′ s, equivalently a′ c′ = b′ d′ . Thus d′ divides a′ c′ and
hence, since c′ and d′ are co-prime, d′ divides a′ . Analogously, since a′ divides b′ d′ and
b′ and a′ are co-prime, a′ divides d′ . It follows that a′ = d′ and a = d. This contradicts
the assumption that all of a, b, c, d are different.
Solution 2. b) Assume for contradiction that such a, b, c, d exist. Write the equation as
a
d
= .
b
c
When we put equal fractions into lowest terms, we get equal fractions (in lowest terms)
with equal numerators and also equal denominators. The number we divide by is the
greatest common divisor of the numerator and the denominator. Since we are given
gcd( a, b) = gcd(c, d) the denominators and numerators will be divided through by the
same number, that is, the numerators and denominators must be equal to begin with.
Thus a = d and b = c, contradicting the assumption that a, b, c, d are different.
OC-3. How many 5-digit natural numbers are there such that after deleting any one
digit, the remaining 4-digit number is divisible by 7? (Juniors.)
Answer: 8.
Solution. Let M = abcde be a number with the required property. By deleting a and b
we get A = bcde and B = acde, respectively. Since they are divisible by 7, so is their
difference B − A = 1000( a − b), hence a − b is divisible by 7, hence a and b are congruent
modulo 7. Analogously, we have that b and c, that c and d, and finally that d and e are
congruent modulo 7. Thus all the digits are congruent modulo 7.
If M has digits that are at least 7, we can subtract 7 from each such digit to obtain a new
number M′ . It is easy to see that M satisfies the condition in the problem if and only if
M′ does. Since all digits give the same remainder, we are left to consider xxxxx where
0 6 x 6 6. By deleting a digit we get xxxx = x · 1111 that is divisible by 7 only if x = 0.
Indeed, 1111 and 7 are co-prime. Thus every digit of M is either 0 or 7. The first two
digits must be 7 (since the number has 5 digits and any number we get by deleting a
digit has 4 digits), the last three digits can be any of 0 or 7 independently. Thus there
are 2 · 2 · 2 = 8 suitable numbers.
OC-4.
A magician wants to do the following trick, using an n-year-old volunteer
from the audience. On a board, the magician writes n different positive integers in a
row. Now, between every two consecutive integers, the volunteer writes the difference
of the inverses of the left-hand and right-hand numbers. He finds that all the differences
are equal. Show that the magician can do the trick with every volunteer who is at least
2 years old. (Juniors.)
Solution. If the volunteer is n years old then the magician can pick a number N that is
divisible by all the integers from 1 to n, e.g. their least common denominator or product,
3
and write on the board the numbers
N N
N
, , ..., .
1 2
n
n
1 2
, , ...,
respectively. We
N N
N
1
see that the differences of consecutive numbers are all equal to − .
N
These are different positive integers whose inverses are
OC-5. A squaric is a square that has been divided into 8 equal triangles by perpendicular bisectors of its sides and its diagonals. Each
of those lines divides the squaric into two parts; we can take one of
the parts and reflect it over a second dividing line that is perpendicular to the original line (equivalently, we rotate along the line by 180◦
in space). Every triangle has been coloured by one of four colours and there are two
triangles of each colour. Show that regardless of the initial colouring, the squaric can be
taken to an end position where at every side of the square both triangles have the same
colour. (Juniors.)
Solution. Denote the positions of triangles by numbers 1 to 8 and axes of reflection by
letters x, u, y, v as seen in Fig. 5. Additionally, let A, B, C, D be the colours used.
y
v
6
u
5
7
4
8
3
1
2
C D
D C
C
D
A A
C B
B C
C
B
⇒
B D
B
B
⇒
B D
A A
x
A A
D C
D
C
Figure 5
A A
D D
C
C
B
B
⇒
C
B
A A
A A
B B
B B
C
C
⇒
D C
B D
⇒
C
B
A A
D D
C
B
⇒
C
B
D D
⇒
C D
A A
D
D
A A
Figure 6
Assume without loss of generality that the triangle in position 1 has colour A. If the
other triangle of colour A is not in position 2, we take it there by reflections that leave
the triangle in position 1 fixed.
Position of colour A triangle Axes of reflection
3
u, y, x
4
u, x
5
x
6
y, x
7
v, y, x
8
v, x
Now the triangles at the bottom side of the squaric have the same colour.
Assume without loss of generality that the triangle at position 3 has colour B. If the
other triangle of colour B is not located at position 4, we take it there as follows, always
leaving the triangle in position 1 unmoved:
4
Position of colour B triangle Axes of reflection
5
y, v, y
6
v, y
7
y
8
v, y, v, y
Now the bottom and right sides of the squaric have triangles of suitable colours.
Finally assume without loss of generality that the triangle in position 5 has colour C.
If the other triangle coloured C lies in position 6, we are done, since then the triangles
at positions 7 and 8 must have colour D. If the other triangle of colour C does not lie
in position 6, we use the following reflections to solve the squaric, always leaving the
triangle in position 1 unmoved (cf. Fig. 6):
Position of colour C triangle Axes of reflection
7
y, u, y, v, y
8
u, y
OC-6. Is it true that every polynomial P( x ) = am x m + . . . + a1 x + a0 with integer
coefficients whose value P(z) for every integer z is a composite number can be written
as P( x ) = Q( x ) · R( x ), where Q and R are polynomials with integer coefficients, neither
of which is constantly 1 or −1? (Seniors.)
Answer: no.
Solution. Let P( x ) = x2 + x + 4. If a is any integer then P( a) = a2 + a + 4 = a( a + 1) + 4.
One of a and a + 1 has to be even, thus P( a) is even. Since a and a + 1 cannot have
opposite signs, a( a + 1) is non-negative, thus P( a) > 4. Thus P( a) is composite.
Write P( x ) = Q( x ) · R( x ), where Q and R are polynomials with integer coefficients.
Since the leading coefficient of P is 1, the leading coefficients of Q and R are both 1 or
both −1. So if Q or R is constant it has to be 1 or −1. If neither is a constant, since P is
a square polynomial, Q and R have to be linear polynomials. But P has no roots, hence
it is not a product of linear polynomials. Therefore P( x ) = x2 + x + 4 cannot be written
as a product of polynomials both different from −1 and 1.
A
OC-7.
Let O be the circumcentre of triangle ABC.
Lines AO and BC intersect at point D. Let S be a point
on line BO such that DS k AB and lines AS and BC intersect at point T. Prove that if O, D, S and T lie on the same
circle, then ABC is an isosceles triangle. (Seniors.)
S
O
Solution. We have that OAB is an isosceles triangle; so is
T
C
D
OSD since DS and AB are parallel (Fig. 7). It follows that B
the triangles OAS and OBD are equal, using that |OA| =
|OB|, |OS| = |OD | and ∠SOA = ∠ DOB. Thus ∠OSA =
Figure 7
∠ODB, from which it follows that ∠OST = ∠ODT. The
points O, D, S, T are located on a circle, so that the points D and S that are symmetric
with respect to the perpendicular bisector of the segment AB are located on different
sides of the line OT. It follows that OST and ODT are opposite angles of the inscribed
quadrilateral ODTS and their sum is 180◦ . Thus ∠OST = ∠ODT = 90◦ . The altitude
5
from vertex A of the triangle ABC goes through the circumcentre of ABC, so it is also
the perpendicular bisector of BC. It is possible only if | AB| = | AC |.
Remark. If ABC is a right triangle, the validity of the claim depends on definitions used.
By the usual high-school definition a right triangle does not satisfy the assumptions of
the problem: if the right angle is at A, then S = D and we cannot talk about line DS; if
the right angle is at B there is no intersection point T; if the right angle is at C, the lines
DS and AB would coincide but coinciding lines are not considered parallel at school.
OC-8. Wolf and Fox play the following game on a board with a finite number of unit
squares. In the beginning, all squares are white and empty. First, Wolf picks up a game
piece from a pile, and either places it on a white square, paints this square gray and
removes all the other pieces from the board, or places it on an empty gray unit square.
Then, Fox makes a move by the same rules, only her colour is red and not gray. The
players continue taking turns and the last player to make a move wins (assume there is
an infinite supply of game pieces). Who wins if both play optimally? (Seniors.)
Answer: Wolf.
Solution 1. Let Wolf have the following strategy. If there are white squares on the board
he will place a piece on one (and colour it gray), otherwise on an empty gray square.
Let us prove this is a winning strategy. Since Wolf starts occupying white squares and
colours them at every move, then after each of his moves there are more gray than red
squares. After every move of Fox there are at least as many gray squares than red. This
is true until there are no more white squares. The square coloured last contains a piece,
all the other squares are empty and the game continues so that each player places pieces
on empty squares.
If the last white square is coloured by Wolf there will be more gray squares than red.
Although the square coloured last contains a piece and he cannot move there anymore,
there are at least as many gray squares left as red. This means Wolf can move after every
move of Fox. If the last square is coloured by Fox, it will contain a piece and she cannot
move there afterwards. Therefore Wolf will have more empty gray squares than Fox has
empty red ones. In both cases Wolf gets to make the last move.
Solution 2. Wolf can, after the first move of each player, colour a white square on his
second move, and after that copy moves of Fox.
Solution 3. It is clear that the game always ends, so somebody has a winning strategy.
Suppose Fox has a winning strategy. At the first move no player has a choice (they
colour a white square). At the second move Wolf has two choices.
• He colours a new white square. Then there are two gray squares, one containing a
piece, and one empty red square on the board. By assumption, Fox has a winning
strategy.
• He places a piece on the gray square. Fox can only colour a new white square. After
Fox moves there are two red squares, one containing a piece and one empty gray
square on the board. By symmetry now Wolf has a winning strategy.
Contradiction.
OC-9.
The teacher gives every student a triple of positive integers. First, every student has to reduce the second and third number by dividing them by their greatest com6
mon divisor, then reduce the first and third number of the resulting triple by dividing
them by their greatest common divisor, and finally, reduce the first and second number
of the new triple by dividing them by their greatest common divisor. Then, everybody
has to multiply the numbers in the final triple and tell the result to the teacher. It is
known that the initial triples only differ by the order of numbers. Find the greatest
possible number of different correct answers that the students could get. (Seniors.)
Answer: 3.
Solution. Let ( a, b, c) be the initial triple and
c
d1 = gcd(b, c),
d2 = gcd a,
,
d1
b
d3 = gcd a,
.
d1
a b
c
b c
and
, ,
, reAfter the first and second division, we get triples a, ,
d1 d1
d2 d1 d1 d2
spectively.
b
c
a b
,
= d3 . Since
and
are co-prime, their divisors d2
Let us prove that gcd
d2 d1
d1
d1
a
b
and d3 are co-prime. Since d3 divides a, it hence divides
. Since d3 divides
, it
d2
d1 a b
a b
a
also divides gcd
,
. On the other hand, since
divides a, clearly gcd
,
d2 d1
d2
d2 d1
divides
d3 whichproves the claim. Therefore the triple after the third division is
b
c
abc
a
,
,
and the correct answer is 2 2 2 .
d2 d3 d1 d3 d1 d2
d1 d2 d3
If we swap b and c in the initial triple, d1 is left unchanged and d2 and d3 are swapped
which leaves the final answer unchanged. Therefore the answer depends only on what
we choose as the first component of the triple. Thus there can not be more than 3 different answers.
We get three different correct answers if we pick a triple ( p2 qr, pq2 r, pqr2 ) where p, q
and r are pairwise different primes. Indeed this triple changes as
( p2 qr, pq2 r, pqr2 ) 7→ ( p2 qr, q, r ) 7→ ( p2 q, q, 1) 7→ ( p2 , 1, 1),
giving the answer p2 . By changing cyclically the order of the components in the triple,
answers q2 and r2 are obtained.
OC-10. In a square grid of dimension m × n where m, n > 5, every square has been
coloured black or white. At each step, we can pick some horizontal or vertical strip of
width 1 and odd length that contains squares of both colours, and colour all squares in
this strip by the colour occurring less in the strip. Show that by these steps we can give
all squares the same colour. (Seniors.)
Solution 1. Let the grid have m rows and n columns. At first we shall show that the
squares in each row can be given the same colour. If n is odd we cover a row by one
strip and colour all the squares by the colour that occurs less in that row. If n is even we
cover all squares in the row except for the first one by a strip (of odd length) and give
them one colour. If the first square is coloured differently from the rest of the row we
7
cover the squares 1, 2 and 3 by a strip and give them the same colour, then we cover the
squares 1, 2, 3, 4 and 5 by a strip and colour them by the colour of squares 4 and 5. Now
all the squares in that row have the same colour.
Let us now do the same construction for columns. After that the squares in each column
have the same colour, but also all the colours are equal since after the first stage all
columns looked identical.
Solution 2. At first, prove that a rectangle with dimensions 1 × 5 that contains squares of
both colours can be coloured by each colour. Now we can make the first row monochromatic by sequentially choosing the 1 × 5 blocks (or not choosing if a block is already
monochromatic). Then we can similarly give every column the colour of its first square.
Remark. One can find other solutions, precisely by dividing the grid into at most four
sub-grids, with each dimension odd and at least 3, and making the sub-grids and eventually the whole grid monochromatic.
OC-11. We are given different positive integers a1 , a2 , . . . , an where n > 3 and every
integer except the first and last one is the harmonic mean of its neighbours. Show that
none of the given integers is less than n − 1. (Seniors.)
1 1
1
form an arithmetic progression. By
,
, ...,
a1 a2
an
1
1
1
>
> . . . > , equivalently a1 < a2 < . . . < an .
symmetry we may assume that
a1
a2
an
Since a1 , a2 , . . . , an are positive integers,
1
1
1
a − a1
n−2
1
1
>
−
−
= ( n − 2) · 2
>
.
= ( n − 2)
a2
a2 a n
a1 a2
a1 a2
a1 a2
Solution. As given, the numbers
1
n−2
>
, and multiplying both sides by a1 a2 gives a1 > n − 2. Since a1 is an
a2
a1 a2
integer, a1 > n − 1. The numbers a2 , . . . , an are greater than a1 and thus greater than
n − 1.
Hence
OC-12. Two circles are drawn inside a parallelogram ABCD so that one circle is tangent to sides AB and AD and the other is tangent to sides CB and CD. The circles touch
each other externally at point K. Prove that K lies on the diagonal AC. (Seniors.)
C
D
O1
A
K
O2
B
Solution 1. Let O1 and O2 be the centres of
the first and the second circle, respectively
(Fig. 8). Consider the triangles O1 AK and
O2 CK. Their angles AO1 K and CO2 K are
equal since their sides O1 K and O2 K lie on the
same line and the sides O1 A and O2 C are par1
1
allel since ∠O1 AB = ∠ DAB = ∠ BCD =
2
2
∠O2 CD. As ∠ DAB = ∠ BCD, we have
Figure 8
|O2 C |
|O1 A|
=
. Therefore the triangles O1 AK and O2 CK are similar. Thus ∠O1 KA =
|O1 K |
|O2 K |
∠O2 KC, from which it follows that the points A, K and C are collinear.
Solution 2. Consider the homothety with centre K that takes one of the circles onto the
8
other one. Then the line AB is taken to the line CD and the line AD is taken to the line
CB. Thus the intersection point A of the lines AB and AD goes to the intersection point
C of the lines CD and CB. It follows that the points A, K and C are collinear.
OC-13.
Let x and y be arbitrary real numbers.
a) If x + y and x + y2 are rational numbers, can we deduce that x and y are rational
numbers?
b) If x + y, x + y2 and x + y3 are rational numbers, can we deduce that x and y are
rational numbers?
(Seniors.)
Answer: a) no; b) yes.
√
√
1+ 2
1− 2
and y =
. Then x + y = 1 and x + y2 =
Solution. a) Pick x =
2
2
that x + y and x + y2 are rational but x and y are not.
b) Let x + y, x + y2 and x + y3 be rational. If y = 0 or y = 1 then both x
rational. If y 6= 0 and y 6= 1 then the following is also a rational:
5
. We see
4
and y are
y3 − y2
y ( y2 − y )
( x + y3 ) − ( x + y2 )
=
=
= y.
( x + y2 ) − ( x + y )
y2 − y
y2 − y
Thus ( x + y) − y = x is also rational.
OC-14.
an =
A sequence ( an ) of natural numbers is given by the following rule:
lcm( an−1 , an−2 )
gcd( an−1 , an−2 )
for all n > 2.
It is known that a560 = 560 and a1600 = 1600. Find all possible values of a2007 . (Seniors.)
Answer: a2007 = 140 is the only possible value.
Solution. Explore the behaviour of the sequence in general. Note that it is sufficient to
consider the behaviour of the sequence for each prime factor separately. We have
lcm( p a , pb )
pmax(a,b)
=
= p| a−b| .
a
b
min
(
a,b
)
gcd( p , p )
p
Therefore for each prime factor one may consider the behaviour of the sequence of exponents. Thus explore the properties of the sequence (bn ) defined by bn = |bn−1 − bn−2 |
for all n > 2. It is easy to see that either all terms of the sequence are even or there
is a cycle (even, odd, odd). Thus bn+3 and bn have the same parity. Also observe that
bn+3 6 bn for all n.
Consider now the prime factors appearing in given terms. For prime factor 7 one has
b560 = 1 and b1600 = 0. Since 3 | 1601 − 560, previous observations imply that b1601 = 1.
Then b1602 = |1 − 0| = 1. Now the fact that 3 | 2007 − 1602 leads to b2007 = 1. Hence the
exponent of 7 in a2007 is equal to 1.
For the prime factor 5, we have b560 = 1 and b1600 = 2. Analogously with previous case
we obtain b1601 = 1 and also b1602 = 1. Hence as before b2007 = 1, thus the exponent of
5 in a2007 is 1.
9
The prime factor 2 remains. For this b560 = 4 and b1600 = 6. Examine the possible
values of b1601 . Since it must have the same parity as b560 and may not be greater than
it, the only candidates are 0, 2 and 4. Suppose b1601 = 0. Then both b1601 and b1600 are
divisible by 6. Taking into account the definition of the sequence (bn ) implies that 6
divides also all previous terms, including b560 . This leads to contradiction, thus b1601 is
not 0. Suppose that b1601 = 2. Since consequent terms are even, all terms of the sequence
must be even. Dividing all terms by 2 leads to sequence, that still satisfies the definition,
thus all previously considered observations must be valid. The term b560 transforms to
2 and the term b1601 to 1, that means they have different parity. This is a contradiction
analogously to the cases of previous primes.
The last remaining possibility is b1601 = 4 (it is easy to see that a corresponding sequence
exists). Now performing calculations we obtain b1602 = 6 − 4 = 2, b1603 = 4 − 2 = 2,
b1604 = 2 − 2 = 0, b1605 = 2 − 0 = 2 and further the cycle (2, 0, 2) repeats, therewith the
value of terms with the number divisible by 3 is 2. Thus b2007 = 2, hence the exponent
of 2 in a2007 is 2.
Since the terms a560 and a1600 have no other prime factors, taking preceding into account
implies that the term a2007 neither has other prime factors. Hence the only solution is
a2007 = 7 · 5 · 22 = 140.
Selected Problems from the Final Round of National
Olympiad
FR-1. On a railway connecting cities A and B, trains run at full speed except for two
railway segments, where poor track conditions force them to slow down. If any one of
those two segments were repaired, the average speed of a train between A and B would
increase by a third. How much would the average speed between A and B increase if
both segments were repaired? (Grade 9.)
Answer: 2 times.
Solution. Let the train journey between A and B take time t when neither segment is
repaired. If the first segment was repaired, the average speed would increase by a third,
4
3
in other words, times, so the journey would take time t. Thus, repairing the first
3
4
1
1
segment would save t time. Similarly, repairing the second segment would save t.
4
4
1
Repairing both segments would save t and the average speed would increase 2 times.
2
FR-2. Find all possible values of abc · ( a + b + c), given that bca = ( a + b + c)3 and
b 6= 0. (Grade 9.)
Answer: 2008.
Solution. There exist five three-digit cubes: 125 = 53 , 216 = 63 , 343 = 73 , 512 = 83 and
729 = 93 . Of these, only 512 satisfies bca = ( a + b + c)3 . Thus, a = 2, b = 5, c = 1 and
abc · ( a + b + c) = 251 · (2 + 5 + 1) = 2008.
10
O3
O2
C
B
B
O1
A
O3
A
O1
C
O2
D
O4
Figure 9
Figure 10
FR-3. a) Circles c1 and c2 touch externally at point A, circles c2 and c3 touch externally
at point B, and circles c3 and c1 touch externally at point C. Suppose that triangle ABC
is equilateral. Are the radii of c1 , c2 and c3 necessarily equal?
b) Circles c1 and c2 touch externally at point A, circles c2 and c3 touch externally at point
B, circles c3 and c4 touch externally at point C, and circles c4 and c1 touch externally at
point D. Suppose that ABCD is a square. Are the radii of c1 , c2 , c3 and c4 necessarily
equal? (Grade 9.)
Answer: a) yes; b) no.
Solution. a) Let O1 , O2 and O3 be the midpoints of c1 , c2 and c3 , respectively (Fig. 9).
Triangles O1 CA, O2 AB and O3 BC are isosceles, as each triangle has two radii of the
same circle as its two sides. Let ∠O1 CA = ∠O1 AC = α, ∠O2 AB = ∠O2 BA = β and
∠O3 BC = ∠O3 CB = γ. Suppose that triangle ABC is equilateral, so ∠ ABC = ∠ BCA =
∠CAB = 60◦ . As ∠O1 AC + ∠CAB + ∠O2 AB = 180◦ , we have α + β = 120◦ . Similarly,
β + γ = 120◦ and γ + α = 120◦ . The last three equations together give α = β = γ = 60◦ .
Thus, triangles O1 CA, O2 AB, O3 BC are equilateral and as ABC is also equilateral, they
are in fact equal.
b) Choose the midpoints of the three circles as O1 (6; 0), O2 (0; 3), O3 (−6; 0), O4 (0; −3)
(Fig. 10). Then O1O2O3O4 is a rhombus and points A(2; 2), B(−2; 2), C (−2; −2),
D (2; −2) on the sides of the rhombus form a square. Take each vertex of the rhombus to be the midpoint of a circle drawn through the two closest vertices of the square.
Then these four circles touch externally at A, B, C, D, yet they do not all have equal
radii (e.g., |O1 A| 6= |O2 A|).
FR-4. Let n be a positive integer. Rays originating from the midpoint X of a revolving stage divide the stage into 2n + 2 equal sectors, coloured alternatingly black and white (n = 3 in the figure).
Similarly, equally spaced rays originating from X divide the fixed
floor area outside the revolving stage into 2n alternatingly blackand-white sectors. Prove that regardless of the position of the revolving stage, there exists a sector of the stage that is completely
embraced by a single fixed floor sector of the same colour. (Grade 9.)
Solution. Consider the rays dividing the revolving stage into 2n + 2 sectors. Since the
rest of the floor is divided into 2n sectors, there exist two neighbouring rays that pass
through the same floor sector. If the revolving stage sector between those two rays has
11
the same colour as the floor sector, we are done. If, on the other hand, the two sectors
are of different colour, then the stage sector symmetrically opposite to the original sector
satisfies our conditions. This stage sector is completely embraced by the floor sector
symmetrically opposite the original floor sector, however, when turning 180◦ , the stage
sectors change colour n + 1 times, while the floor sectors change colour only n times, so
the two sectors symmetrically opposite to the original sectors are of the same colour.
FR-5. Circles c1 and c2 with midpoints O1 and O2 intersect at point P. Circle c2 intersects O1O2 at point A. Prove that there exists a circle touching c1 at P and O1O2 at A iff
∠O1 PO2 = 90◦ . (Grade 10.)
Solution. Assume first there exists a circle c touching c1 at P and O1O2 at A (Fig. 11). Let
O be the midpoint of c, then line O1O passes through P. Consider triangles OPO2 and
OAO2 . Clearly |OP| = |OA| and |O2 P| = |O2 A| as the radii of circles c and c2 . Also,
the triangles share a third side OO2 , so they are equal. As ∠OAO1 = 90◦ , we must also
have ∠OPO2 = 90◦ .
Assume now ∠O1 PO2 = 90◦ . Then line O2 P is perpendicular to radius O1 P and thus
touches c1 at P. As |O2 P| = |O2 A|, the line drawn through P perpendicular to O2 P
and the line drawn through A perpendicular to O2 A intersect at a point O such that
|OP| = |OA|. A circle with midpoint O and radius OP then touches c1 at P and O1O2
at A.
P
O
O1
A
C
O2
A
D
B
D′
Figure 11
Figure 12
FR-6. Do there exist 5 different points in the plane such that all triangles with vertices
at these points are right triangles and
a) no four of the chosen points lie on the same line;
b) no three of the chosen points lie on the same line?
(Grade 10.)
Answer: a) yes; b) no.
Solution 1. a) Choose four vertices of a square and the intersection point of its diagonals.
b) Consider a set of points in the plane such that all triangles with vertices in those
points are right triangles and no three points lie on the same line. Choose some two
points A and B; all the remaining points then lie either on the circle with diameter AB,
or on either line perpendicular with AB drawn through endpoint A or B (Fig. 12). At
most four points (including A and B) can lie on the circle, since any two of such three
12
points must be the two endpoints of some diameter. Also, in addition to A and B, there
can be at most one point on either perpendicular.
Suppose now that C and D are two points satisfying our conditions such that C lies on
the circle and D lies on one of the two lines, say, on the perpendicular drawn through
B. If C and D lie on the same side of AB, then ∠ ACD > ∠ ACB = 90◦ , and ACD is
not a right triangle. If, on the other hand, C and D lie on opposite sides of AB, then
∠ DBC > ∠ DBA = 90◦ , so DBC is not a right triangle. Thus, either all points lie on the
circle, or they all lie on the two perpendiculars. In either case, there can be at most 4
such points.
Solution 2. b) Assume by contradiction that it is possible to choose 5 points satisfying
the conditions. Since each three points form the vertices of a right triangle, there are 10
right triangles with vertices in these 5 points. Thus, there exists a point O that is the
vertex of at least two right angles. Let OAB and OXY be the two triangles with right
angles at O.
Now, if either X or Y was lying on line OA, the other point would have to lie on AB.
But then we would have three points on the same line, since at most one of X and Y
can coincide with A or B. Analogously, neither X nor Y can lie on OB. Now if X (resp.
Y) and B lie on opposite sides of line OA, then XOB (resp. YOB) is an obtuse triangle.
Similarly, X and A (or Y and A) cannot lie on opposite sides of OB. Thus, both X and Y
must lie within the right angle AOB, but then XOY is not a right triangle.
FR-7. Call a rectangle splittable if it can be divided into two or more square parts such
that the side of each square is of integral length and there is a unique square with smallest side length. Find the dimensions of the splittable rectangle with the least possible
area. (Grade 11.)
Answer: 5 × 7.
Solution. The unique smallest square of the partition cannot lie
on the side of the rectangle, for it would have a larger square
on either side and the area between the two squares could only
be filled by squares no larger than the smallest square. Analogously, the smallest square cannot lie in the corner. Now, the
distance between the smallest square and any side of the triangle
must be at least one unit longer than the side length of the smallFigure 13
est square, for otherwise the area between the smallest square
and the side could not be filled. Thus, the length of each side of the rectangle is at least
1 + 2 + 2 = 5 and each square on a side must have side length at least 2. Thus, if the
rectangle has a side of length 5, on this side we must have a square with side length at
least 3. But then the distance between the smallest square and this side is at least 3. The
same holds for the opposite side of length 5, so the length of the longer side must be at
least 1 + 3 + 3 = 7. It is possible to partition a 5 × 7 rectangle in the desired way (see
Fig. 13). The area of this rectangle is 35, which is indeed the smallest possible area, since
any rectangle with shorter side length greater than 5 has area at least 6 · 6 = 36.
FR-8. Circles c1 and c2 with respective diameters AB and CD of different length touch
externally at point K. An external tangent common to both circles touches c1 at A and
c2 at C. Line BD intersects c1 again at point L and c2 at point M. Prove that triangles
13
C
A
C
K
O1
B
A
O1
O2
D
L
M
L
Figure 14
B
K
M
O2
D
Figure 15
AKL and BKM are similar. (Grade 11.)
Solution. Let O1 and O2 be the midpoints of circles c1 and c2 , respectively (Fig.
14 and 15). The isosceles triangles BO1 K and CO2 K are similar, since their corresponding legs are parallel: AB k CD and point K lies on O1O2 . Thus, the bases are also
parallel, so K lies on BC. Now on one hand, ∠KAL = ∠KBM, while on the other hand
∠ ALK = ∠ ABK = ∠KCD = ∠KMB. We see that triangles AKL and BKM have two
pairs of equal angles and hence are indeed similar.
FR-9. Let a, b, c be real numbers. Prove that a2 + 4b2 + 8c2 > 3ab + 4bc + 2ca. When
does equality hold? (Grade 11.)
Answer: Equality holds iff a = 2b = 4c.
Solution. Bringing all terms to the lhs, we get
a2 + 4b2+ 8c2 − 3ab − 4bc− 2ca =
3 2
1 2
2
2
2
2
=
a − 3ab + 3b + b − 4bc + 4c + 4c − 2ca + a =
4
4
!2
!2
√
√
1
3
a − 3b + (b − 2c)2 + 2c − a
> 0.
=
2
2
Equality holds iff equations
√
√
1
3
a = 3b, b = 2c, 2c = a hold simultaneously, in other
2
2
words, iff a = 2b = 4c.
Remark. One may find other solutions, precisely using AM-GM on (1.5a2 , 6b2 ),
!2
!2
√
√
7
3
(0.5a2 , 8c2 ), (2b2 , 8c2 ), grouping the lhs as a − b − c +
b − 7c , or con2
2
sidering the lhs as a quadratic trinomial in a, b and c and investigating the respective
discriminants.
FR-10. Does there exist a convex hexagon ABCDEF such that the circumcircles of
triangles ABC, CDE and EFA intersect at a common point inside the hexagon? (Grade
11.)
Answer: no.
14
B
Solution. Suppose that such a hexagon exists and let O be
the common intersection point of the three circumcircles
(Fig. 16). Then quadrilaterals ABCO, CDEO and EFAO are
all inscribed, so ∠ BAO + ∠ BCO = 180◦ , ∠ DCO + ∠ DEO =
180◦ and ∠ FEO + ∠ FAO = 180◦ . Adding the three equations, we get ∠ BCD + ∠ DEF + ∠ FAB = 3 · 180◦ . On the
other hand, all internal angles of a convex hexagon are
less than 180◦ , so the sum of the three angles cannot reach
3 · 180◦ , contradiction.
FR-11. Find the least possible value of (1 + u2 )(1 + v2 ),
where u and v are real numbers satisfying u + v = 1. (Grade
12.)
Answer:
A
O
C
F
D
E
Figure 16
25
.
16
1
1
+ x and v = − x. Then
2
2
1
2 2 1
1
2
2
2
(1 + u )(1 + v ) = 1 +
+x
−x
1+
= 1+ +x +x ·
2
2
4
2
5
25 3 2
25 5 2
1
+ x2 − x2 =
+ x + x4 − x2 =
+ x + x4 .
· 1 + + x2 − x =
4
4
16 2
16 2
Solution 1. Write u =
3
Since x2 and x4 are both non-negative, the obtained sum is minimal when x = 0. The
2
25
latter gives (1 + u2 )(1 + v2 ) = .
16
Solution 2. As u + v = 1, we get
(1 + u2 )(1 + v2 ) = 1 + u2 + v2 + u2 v2 =
= 1 + (u + v)2 − 2uv + u2 v2 = 2 − 2(uv) + (uv)2 .
Let s = uv. For a fixed sum u + v = 1, the product s = uv is maximal when u = v. Thus,
1 2
1
= . Now, we need to minimize 2 − 2s + s2 = (s − 1)2 + 1,
we can bound s 6
2
4i
1
1
which is decreasing in −∞;
and obtains the minimum at s = .
4
4
Solution 3. Notice that u = 1, v = 0 gives (1 + u2 )(1 + v2 ) = 2, while for any u > 1
or v > 1 (or equivalently, v < 0 or u < 0), (1 + u2 )(1 + v2 ) > 2. Thus, we may
restrict to the case u, v ∈ [0, 1]. Now consider a triangle ABC such that its side BC and
altitude AH (Fig. 17) have unit length and H divides BC to parts of length u and v. Then
1
1
1
u + v = 1 and the law of sines gives · | AB| · | AC | · sin ∠ BAC = · | BC | · | AH | = ,
2
2
2
1
. The value sin ∠ BAC is maximal when
so (1 + u2 )(1 + v2 ) = | AB|2 | AC |2 =
sin2 ∠ BAC
H is the midpoint of BC. Indeed, let c be the circumcircle of ABC in the case when
H is the midpoint. For any other configuration, A lies outside this circle c and thus
the angle BAC is smaller (note that the angle BAC is always acute as BC cannot be the
5
longest side of ABC). Now if H is the midpoint of BC, we get | AB|2 = | BC |2 = , and
4
15
25
.
16
Remark. One may find other solutions, precisely determining the minima of g(u) =
(1 + u2 )(1 + (1 − u)2 ) using derivatives, or writing out Jensen’s inequality for a convex
function l ( x ) = ln(1 + x2 ).
| AB|2 | BC |2 =
A
A′
A
D
O1
O
B
B
u
H
v
C
O2
E
C
Figure 17
Figure 18
FR-12. In a convex quadrilateral ABCD, | AB| = | BC | = |CD |. Diagonals AC and
BD intersect at point O. Prove that the circumcircles of triangles AOB and COD are
mutually tangent iff AC is perpendicular to BD. (Grade 12.)
Solution. Assume first that the circumcircles of AOB and COD are mutually tangent
(Fig. 18). Draw a tangent common to both circles from O, and let the tangent line intersect BC at E. Then | AB| = | BC | implies ∠EOB = ∠OAB = ∠ BCO. Similarly, ∠EOC =
∠ODC = ∠CBO. Triangle BOC now gives ∠EOB + ∠EOC + ∠OBC + ∠OCB = 180◦ or
2∠EOB + 2∠EOC = 180◦ , so finally ∠ BOC = ∠EOB + ∠EOC = 90◦ .
Assume now AC is perpendicular to BD. The circumcentres O1 and O2 of right triangles
AOB and COD lie on the respective hypotenuses AB and CD. We have ∠O1OA =
∠O1 AO = ∠ BCO and ∠O2OD = ∠O2 DO = ∠CBO. As BOC is also a right triangle,
∠ BCO + ∠CBO = 90◦ . Finally, ∠O1OA + ∠ AOD + ∠O2OD = ∠ BCO + 90◦ + ∠COB =
180◦ , so the circumcircles of AOB and COD touch at O.
FR-13. All natural numbers that are less than a fixed positive integer n and relatively
prime to it are added one-by-one in increasing order. How many intermediate sums
(starting from the lonely first addend and including the final sum) are divisible by n, if
a) n is an odd prime number?
b) n is the square of an odd prime number?
(Grade 12.)
Answer: a) 1; b) 1.
Solution. a) Let n = p where p is an odd prime. The addends are 1, 2, . . . , p − 1, thus the
intermediate sums have form 1 + . . . + k where 1 6 k 6 p − 1. Suppose p | 1 + . . . + k.
k ( k + 1)
Then p | k (k + 1) as 1 + . . . + k =
. Thus either p | k or p | k + 1. This is possible
2
only for k = p − 1.
b) Let n = p2 where p is an odd prime. Any number less than p2 is added if and only
if it is not divisible by p. Divide the addends into p groups, each consisting of p − 1
16
members:
1,
2, . . . . . . ,
p − 1,
p + 1,
p + 2, . . . . . . ,
p + p − 1,
.......................................................
( p − 1) p + 1, ( p − 1) p + 2, . . . . . . , ( p − 1) p + p − 1.
Let an intermediate sum be divisible by p2 ; then it is divisible by p, too. As the rows
are equivalent modulo p, we can use part a) of the problem to deduce that the last
intermediate sum of every row is divisible by p and the others are not. Hence, in the
whole intermediate sum under consideration, the last row cannot occur partially, i.e.,
our intermediate sum consists of whole rows of addends.
p ( p − 1)
The sum of the elements of the first row is
. The sum of the numbers of each
2
following row is by p( p − 1) larger than that of the row preceding it. Thus the row sums
p ( p − 1)
p ( p − 1)
p ( p − 1)
,3·
,5·
etc.. The sum of the numbers of the first i rows
are 1 ·
2
2
2
p ( p − 1)
p ( p − 1)
is (1 + 3 + . . . + (2i − 1)) ·
= i2 ·
.
2
2
p ( p − 1)
then p | i2 · ( p − 1), implying p | i. Hence i = p, i.e., the sum is the
If p2 | i2 ·
2
final sum.
Remark. It is easy to show that the entire sum of ϕ(n) addends is divisible by n for all
integers n > 2. If n is neither a prime nor the square of a prime then there can be more
intermediate sums divisible by n. For example, if n = 16 then the intermediate sum
1 + 3 + 5 + 7 containing only half of the addends is divisible by 16. If n = 27 or n = 39
then two intermediate sums in addition to the final sum are divisible by n, etc.
FR-14. Consider a point X on line l and a point A outside the line. Prove that if there
exists a point Z1 on l such that the three side lengths of triangle AXZ1 are all rational,
then there exist two other points Z2 and Z3 on l such that the side lengths of triangles
AXZ2 and AXZ3 are also all rational. (Grade 12.)
Solution. We consider three separate cases.
• If AXZ1 is equilateral, i.e., | AX | = | AZ1 | and ∠XAZ1 = 60◦ , then take Z2 on
the extension of Z1 X across X such that | XZ2 | = 0.6 | AX |, and take Z3 to be the
reflection of Z2 across the perpendicular bisector of XZ1 (Fig. 19). The law of
cosines implies | AZ2 | = | AX |2 + | XZ2 |2 − 2 · | AX | · | XZ2 | · cos 120◦ = 1.96 | AX |2 ,
so | AZ2 | = 1.4 | AX | and the side lengths of AXZ2 as well as AXZ3 are rational.
Let now AXZ1 be not equilateral.
A
A
A
Z2
X
Z1
Figure 19
Z3
Z3
X
Z1 Z2
Figure 20
17
X
Z3
Figure 21
Z1
Z2
• Assume | XZ1 | 6= | AX | and | XZ1 | 6= | AZ1 |. Choose Z2 on ray XZ1 such that
∠Z2 AX = ∠ AZ1 X, and choose Z3 on ray Z1 X such that ∠Z3 AZ1 = ∠ AXZ1
(Fig. 20). Points Z2 and Z1 differ since ∠ AZ1 X 6= ∠Z1 AX, and points Z3 and X
differ since ∠ AXZ1 6= ∠XAZ1 . Triangles Z2 XA and Z3 AZ1 are similar to triangle
| AX |
| Z A|
AXZ1 with similarity ratios
and 1 , so their side lengths are rational, and
| Z1 X |
| Z1 X |
| XZ3 | is rational, too.
• Assume w.l.o.g. | XZ1 | = | AZ1 | (Fig. 21). Choose Z2 as before, then Z2 differs from
X and Z1 and the side lengths of AXZ2 are rational. Take Z3 to be the reflection of
Z1 across the perpendicular bisector of XZ2 . Then Z3 differs from Z1 , as AXZ1 is
not an isosceles right triangle. Triangle AXZ3 is equal to triangle AZ2 Z1 , and the
latter has rational side lengths.
FR-15. A finite number of thin straight pins are attached to a vertical wall such that
no two pins touch each other. If a pin is detached, it slides straight down the wall,
keeping its original angle to the floor. Prove that there exists a pin that can slide freely
down to the floor without being stopped by any of the other pins. (Grade 12.)
Solution 1. If there exists a vertical pin that can slide freely, we are done. Assume now
that no vertical pin can slide down freely. Draw a horizontal line l where the wall meets
the floor and project the endpoints of each pin onto l. If there are no pin points between
some left endpoint and l, colour the projection point on l blue. Similarly, if some right
endpoint is the lowermost pin point on its projection line, colour the corresponding
projection point yellow. Clearly, the leftmost projection point on l is coloured blue, while
the rightmost point is yellow. Thus, moving on l left-to-right, some two consecutive
coloured points must be blue and yellow, respectively. We claim that these points are
the two endpoints of the same pin, and thus this pin can slide down. Indeed, on the
segment between the blue and the yellow point, any lowest pin point above l must
belong to the same pin as the left (blue) and the right (yellow) endpoint.
Solution 2. We prove by induction on the number of pins. The claim clearly holds for
one pin. Assume there is more than one pin, and consider three cases.
1. There exists a pin p such that below every point of p, there is a point of some other
pin. Remove p, then by the induction assumption, some pin v can slide down freely.
Now, put p back. Then p cannot be the only pin stopping v, since below every point
of p, there is a point of another pin, and at least one of those pin points should also
be stopping v.
2. There exists a pin p which cannot slide down freely such that all pins stopping p
lie entirely below p. Remove p and all the remaining pins that do not lie below
p. Then, there must exist a pin v that can slide down freely, but then v can also
slide down in the original configuration, since the only pins possibly stopping it
are those below p.
3. If the previous two cases do not hold, then each pin has some points that have no
other pin points below them, and either the pin can slide down or one of the pins
stopping it does not lie entirely below this pin. Let p be the pin with the rightmost
point amongst all pins (if there is more than one such pin, choose the one with the
topmost such point). Remove p. By the induction assumption, there now exists a
18
pin v that can slide down. Put p back. If v can still slide down, we are done. In the
other case, the only pin stopping it is p. Since v must have some free points with
no other pin points below, the left endpoint of v must reach further left than the left
endpoint of p. We claim that now p can slide down. Indeed, any pin points below
p that also lie below v cannot be stopping p, as they would also be stopping v. But
any other pin below p can also not be stopping p, as p has the rightmost endpoint,
so any pins stopping p should lie completely below p.
Remark 1. One can find other solutions, precisely using a directed graph with pins as
its vertices and an edge from vertex a to vertex b if pin b is stopping pin a: it suffices to
prove that this graph does not contain any directed cycles.
Remark 2. The claim does not always hold when the pins are not straight. For example,
two half-circle pins can be placed to mutually stop each other.
IMO team selection contest
First day
TS-1. There are 2008 participants in a programming competition. In every round,
all programmers are divided into two equal-sized teams. Find the minimal number of
rounds after which there can be a situation in which every two programmers have been
in different teams at least once.
Answer: 11.
Solution 1. After every round consider the biggest set of programmers where the programmers have been in the same team in all rounds so far. Before the first round it
consists of 2008 programmers. With every round its size can decrease by at most twice,
since the programmers belonging to it are divided among two teams in the new round
and at least half of them will again be in the same team. Thus the number of rounds is
at least log2 2008, i.e. at least 11.
We shall show that 11 rounds suffice. Order the 2008 programmers in some way and
add both at the end and at the beginning 20 imaginary programmers. Number the programmers by 11-digit binary numbers from 0 to 2047, adding leading zeros if necessary.
In round i the programmers are divided into teams according to the ith digit of their
number. In every round the kth imaginary programmer from the beginning and the kth
imaginary programmer from the end are in different teams since their corresponding binary numbers have all digits different. Hence both teams have in every round an equal
number of programmers. Also, every pair of programmers belong to different teams in
at least one round since their numbers differ in at least one binary digit.
Solution 2. Let us prove by induction on k that if the number of programmers 2n satisfies the inequalities 2k−1 < 2n 6 2k then k rounds suffice. If k = 1 then we have 2
programmers and clearly one round is enough. Assume the claim is true for some k.
Assume there are 2n programmers where 2k < 2n 6 2k+1 . Divide them into two groups
of s = 2k and t = 2n − s programmers, and number them by 1, . . . , s and s + 1, . . . , s + t
respectively. By the induction hypothesis the programmers in both the first and the second group can be divided into equal-sized teams so that after k rounds every two (in
19
each group) have competed against each other at least once. For rounds 1, . . . , k make
up two new equal teams by taking one team corresponding to each group and putting
them together. Assume without loss of generality that in round k one team consists of
s
the programmers of the first group with numbers 1, . . . , and of the second group with
2
t
numbers s + 1, . . . , s + . In the new round swap and make up a team of programmers
2
t
s
with numbers 1, . . . , and s + + 1, . . . , s + t. We can check that every two program2
2
mers (also from different groups) have now been in different teams at least once.
The other part can be done like in Solution 1.
TS-2. Let ABCD be a cyclic quadrangle whose midpoints of diagonals AC and BD
are F and G, respectively.
a) Prove the following implication: if the bisectors of angles at B and D of the quadq
1
rangle intersect at diagonal AC then · | AC | · | BD | = | AG | · | BF | · |CG | · | DF |.
4
b) Does the converse implication also always hold?
Answer: b) No.
Solution 1. a) Let E be the intersection point of the bisectors from B and D. By the
bisector property,
| AB|
| AE|
| AD |
=
=
.
| BC |
| EC |
| DC |
(1)
By Ptolemy’s theorem, | AB| · |CD | + | AD | · | BC | = | AC | · | BD |. Using this in (1), we
obtain
2 · | BC | · | AD | = | AC | · | BD |,
(2)
2 · | AB| · |CD | = | AC | · | BD |.
(3)
Let F be the midpoint of AC. Then ∠ FAD = ∠CAD = ∠CBD. By (2),
| AC |
| FA|
=
=
| AD |
2| AD |
| BC |
. Hence triangles FAD and CBD are similar. Analogously by (3), triangles FAB and
| BD |
| FA|
| FB|
CDB are similar. Consequently, triangles FAD and FBA are similar. Thus
=
| FD |
| FA|
which implies
1
| AC |2 = | FB| · | FD |.
4
(4)
| DA|
| DC |
=
. Thus bisector property implies that the bisectors of angles at A
| AB|
|CB|
and C intersect at diagonal BD. Let G be the midpoint of BD. Analogously to what we
did before, we obtain
By (1),
1
| BD |2 = | AG | · |CG |.
4
(5)
20
The desired claim follows now by multiplying the corresponding sides of (4) and (5)
and finding the square root.
b) Let ABCD be a rectangle where | AB| > | BC |. Clearly | AG | = | BF | = |CG | = | DF | =
1
1
| AC | = | BD |, implying the rhs of the implication of part a) (lhs of the converse). But
2
2
| AB|
| AD |
> 1 >
shows that the bisectors of angles at B and D do not intersect on
| BC |
| DC |
diagonal AC. Hence the converse implication is false.
Solution 2. a) Denote the interior angles of ABCD by ∠ A, ∠ B, ∠C, ∠ D. In triangle DAB,
cosine law gives
| BD |2 = | AB|2 + | AD |2 − 2 · | AB| · | AD | · cos ∠ A.
In triangle BCD, taking into account that ∠C = 180◦ − ∠ A, cosine law gives
| BD |2 = |CB|2 + |CD |2 + 2 · |CB| · |CD | · cos ∠ A.
Multiplying these two equalities leads to
| BD |4 = (| AB|2 + | AD |2 )(|CB|2 + |CD |2 ) − 4 · | AB| · | AD | · |CB| · |CD | · cos2 ∠ A +
+ 2 (| AB|2 + | AD |2 ) · |CB| · |CD | − (|CB|2 + |CD |2 ) · | AB| · | AD | · cos ∠ A.
−→
−→ −→
On the other hand, 2 AG = ( AB + AD ) implies
4 · | AG |2 = | AB|2 + | AD |2 + 2 · | AB| · | AD | · cos ∠ A
and, analogously (using ∠C = 180◦ − ∠ A),
4 · |CG |2 = |CB|2 + |CD |2 − 2 · |CB| · |CD | · cos ∠ A.
Multiplying these equalities leads to
16 · | AG |2 · |CG |2 =
= (| AB|2 + | AD |2 )(|CB|2 + |CD |2 ) − 4 · | AB| · | AD | · |CB| · |CD | · cos2 ∠ A −
− 2 (| AB|2 + | AD |2 ) · |CB| · |CD | − (|CB|2 + |CD |2 ) · | AB| · | AD | · cos ∠ A.
If the bisectors of angles by B and D intersect on diagonal AC, the bisector property
| AD |
| AB|
=
or | AB| · |CD | = | AD | · |CB|. Thus
gives
|CB|
|CD |
(| AB|2 + | AD |2 ) · |CB| · |CD | − (|CB|2 + |CD |2 ) · | AB| · | AD | =
= | AB| · | AD | · |CB|2 + | AD | · | AB| · |CD |2 − (|CB|2 + |CD |2 ) · | AB| · | AD | = 0.
Consequently, | BD |4 = 16 · | AG |2 · |CG |2 . Considering triangles ABC and CDA, we
obtain in a similar way that | AC |4 = 16 · | BF |2 · | DF |2 . Multiplying the last equalities
and taking the 4th root from both, we obtain the desired result.
21
TS-3. Let n be a positive integer and x, y positive real numbers such that x n + yn = 1.
Prove the inequality
!
!
n
n
1 + x2k
1 + y2k
1
∑ 1 + x4k ∑ 1 + y4k < (1 − x)(1 − y) .
k =1
k =1
1
1 + x2k
<
. Indeed,
Solution. Note first that
1 + x4k
xk
1 + x2k
1
x k + x3k − 1 − x4k
( x3k − 1)(1 − x )
−
=
=
< 0,
1 + x4k
xk
(1 + x4k ) x k
(1 + x4k ) x k
since the conditions of the problem imply 0 < x < 1. Now we estimate
n
1
xn − 1
1 + x2k
< ∑ k = n
.
∑
4k
x ( x − 1)
k =1 x
k =1 1 + x
n
A similar inequality can be proven for y, so we obtain
!
!
!
!
n
n
n
n
1 + x2k
1 + y2k
1
1
∑ 1 + x4k ∑ 1 + y4k < ∑ xk ∑ yk =
k =1
k =1
k =1
k =1
=
yn − 1
1
xn − 1
·
=
.
x n ( x − 1) y n ( y − 1)
( x − 1)(y − 1)
Remark. This problem, proposed by Estonia, appeared in the IMO-2007 Shortlist.
Second day
TS-4.
Sequence ( Gn ) is defined by G0 = 0, G1 = 1 and Gn = Gn−1 + Gn−2 + 1 for
every n > 2. Prove that for every positive integer m there exist two consecutive terms
in the sequence that are both divisible by m.
Solution. Define G−1 = 0, then Gn = Gn−1 + Gn−2 + 1 holds also when n = 1. Consider
the pairs ( Gn , Gn+1 ) of consecutive members of the sequence. There are only m2 pairs
modulo m, hence there are pairs ( Gk , Gk+1 ) and ( Gl , Gl +1 ) with k < l that are componentwise congruent modulo m. Since Gn−2 = Gn − Gn−1 − 1, two consecutive terms
in the sequence determine the previous term uniquely. The same is true modulo m.
Therefore also pairs ( Gk−1 , Gk ) and ( Gl −1 , Gl ) are componentwise congruent modulo
m. Continuing, we see that ( G−1 , G0 ) and ( Gl −k−1 , Gl −k ) are componentwise congruent modulo m. Since G−1 = G0 = 0, the terms Gl −k−1 and Gl −k are divisible by m as
required.
TS-5.
Points A and B are fixed on a circle c1 . Circle c2 , whose centre lies on c1 ,
touches line AB at B. Another line through A intersects c2 at points D and E, where D
lies between A and E. Line BD intersects c1 again at F. Prove that line EB is tangent to
c1 if and only if D is the midpoint of the segment BF.
22
Solution 1. Let K be the second intersection point of the line AD and the circle c1 (Fig. 22).
The triangles KFD and BAD are similar since the corresponding angles are equal. The
triangle BAD is similar to the triangle EAB since, by tangent-secant theorem, ∠ ABD =
∠ BED and they have a common angle at the vertex A. Let O be the centre of the circle
c2 . Since AB is the tangent to the circle c2 at point B, AB ⊥ BO. It follows that AO is a
diameter of the circle c1 since O is on the circle c1 by assumption. Hence also OK ⊥ AK
from which it follows that OK is an altitude of the isosceles triangle ODE. Thus | DK | =
|KE|.
A
The line EB is tangent to the circle c1 at B if and only if ∠EBK =
∠ BAD. Since ∠ ABD = ∠ BED, the last equality is equivalent to F
the triangles EKB and BAD being similar. By the same equality
| AB|
D
=
of angles, the two triangles are similar if and only if
B
| BE|
| DB|
. Since EAB and KFD are similar triangles, the last equality
|KE|
K
O
| DB|
| FD |
=
. This is equivalent to | FD | =
is equivalent to
| DK |
|KE|
| DB| since the denominators are equal.
E
Solution 2. As in the first solution we show that | DK | = |KE|.
Let | AD | = x, | DK | = |KE| = y, | BE| = z, | DB| = u, | FD | = v,
Figure 22
| AB| = w. By the property of intersecting chords, uv = xy.
2
Since AB is a tangent, w = x ( x + 2y). The triangles ABD and
AEB are similar since ∠ ABD = ∠ BED and at vertex A they
u
z
uw
have a common angle. Thus = and hence z =
.
x
w
x
The condition that the line EB is tangent to the circle c1 is equivalent to
z2 = y( x + 2y). We shall show that the last condition is equivalent to u = v:
u2 w2
= y( x + 2y) ⇔ u2 x ( x + 2y) = x2 y( x + 2y) ⇔
x2
⇔ u2 = xy ⇔ u2 = uv ⇔ u = v.
z2 = y( x + 2y) ⇔
TS-6.
A string of parentheses is any word that can be composed by the following rules.
1) () is a string of parentheses.
2) If s is a string of parentheses then (s) is a string of parentheses.
3) If s and t are strings of parentheses then st is a string of parentheses.
The midcode of a string of parentheses is the tuple of natural numbers obtained by finding, for all pairs of opening and its corresponding closing parenthesis, the number of
characters remaining to the left from the medium position between these parentheses,
and writing all these numbers in non-decreasing order. For example, the midcode of
(()) is (2, 2) and the midcode of ()() is (1, 3). Prove that midcodes of arbitrary two
different strings of parentheses are different.
Solution. We can assume that the two strings have equal lengths because otherwise their
midcodes differ by length. We prove the desired claim by induction on the length. In the
23
case of length 2, the claim holds trivially. Let s and t be two longer strings of parentheses.
Consider, for both of them, the longest prefix that forms a string of parentheses itself.
The first and the last character of such prefix form a pair of opening and corresponding
closing parenthesis.
If the prefixes of s and t under consideration have different lengths 2k and 2l, respectively, where assume w.l.o.g. that k < l, then consider the first k numbers in the midcodes of both strings. Let the opening parentheses occur at positions a1 , . . . , ak and the
corresponding closing parentheses occur at positions b1 , . . . , bk in word s. The number
a + bi − 1
. As the first 2k
in midcode that corresponds to the ith pair of parentheses is i
2
characters of s form a string of parentheses, numbers a1 , . . . , ak , b1 , . . . , bk are precisely
1, . . . , 2k in some order. Thus the sum of k smallest members of the midcode of s is
k
1 + 2 + . . . + 2k k
a i + bi − 1
=
− .
2
2
2
i =1
∑
In the midcode of t, the sum of k smallest members is larger since, otherwise, the sum
of position indices of some k pairs of parentheses would be 1 + 2 + . . . + 2k. This would
imply that the corresponding closing parenthesis for each opening parenthesis among
those at positions 1, 2, . . . , 2k occurs within the same positions, leading to l 6 k, a contradiction.
If both prefixes under consideration have length 2k then, for both cases, the part of the
word between the first and the last character of the prefix forms a string of parentheses,
as does the part of the word remaining after the prefix (provided they are non-empty).
As s and t differ, either the first mentioned parts of the words or the second mentioned
parts differ.
In the former case, the induction hypotheses implies that their midcodes also differ. In
the midcodes of s and t, these midcodes are represented by numbers that are by 1 larger,
whereby all these numbers are less than 2k. In addition, both midcodes contain k (from
the pair of parentheses embracing the prefix) and the remaining numbers are larger than
2k. Thus the midcodes of s and t differ.
In the latter case, the induction hypothesis again implies that the midcodes of the parts
after the prefix are different. In the midcodes of s and t, these midcodes are represented
by numbers that are by 2k larger. All other numbers in the midcode are less than 2k.
Hence the midcodes differ.
Remark. A tuple of positive integers x1 , . . . , xn is a midcode of some string of parentheses
k
iff it is monotone,
∑ xi > k
2
n
for every k = 1, . . . , n, and
i =1
∑ xi = n2 .
i =1
Problems listed by topics
Number theory: OC-2, OC-3, OC-4, OC-9, OC-14, FR-13, TS-4
Algebra: OC-6, OC-11, OC-13, FR-1, FR-2, FR-9, FR-11, TS-3
Geometry: OC-7, OC-12, FR-3, FR-5, FR-8, FR-10, FR-12, FR-14, TS-2, TS-5
Discrete mathematics: OC-1, OC-5, OC-8, OC-10, FR-4, FR-6, FR-7, FR-15, TS-1, TS-6
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